Answer:
(a) 2.75 fm
(b) 2.89 fm
(c) 4.70 fm
(d) 7.12 fm
Explanation:
For a given element, the radius r of its nuclei is given by;
r = r₀
Where;
A = Atomic mass of the element
r₀ = 1.2 x 10⁻¹⁵m = 1.2fm
Now let's solve for the given elements
(a) ¹²₆C
Carbon element => This has an atomic mass number of 12
Therefore its radius is given by;
r = 1.2 x 
r = 1.2 x 2.29
r = 2.75 fm
(b) ¹⁴₇N
Nitrogen element => This has an atomic mass number of 14
Therefore its radius is given by;
r = 1.2 x 
r = 1.2 x 2.41
r = 2.89 fm
(c) ⁶⁰₂₇Co
Cobalt element => This has an atomic mass number of 60
Therefore its radius is given by;
r = 1.2 x 
r = 1.2 x 3.92
r = 4.70 fm
(d) ²⁰⁸₈₂Pb
Lead element => This has an atomic mass number of 208
Therefore its radius is given by;
r = 1.2 x 
r = 1.2 x 5.93
r = 7.12 fm
Answer:
The magnitude and direction of electric field midway between these two charges is 
along AB.
Explanation:
Given that,
First charge 
second charge 
Distance = 20 cm
We need to calculate the electric field
For first charge,
Using formula of electric field
Put the valueinto the formula
Direction of electric field along AB
We need to calculate the electric field
For second charge,
Using formula of electric field
Put the valueinto the formula
Direction of electric field along AO
We need to calculate the net electric field at midpoint
Direction of net electric field along AB
Hence, The magnitude and direction of electric field midway between these two charges is along AB.
Answer:
ω = 0.571 rad/s
Explanation:
given data
radius = 30 m
solution
we take here g = 9.8 m/s²
and g is express as
g = r × ω² ....................1
put here value and we get
9.8 = 30 × ω²
solve it we get
ω = 0.571 rad/s
“a point representing the mean position of the matter in a body or system.”
